What is the sum from n=9 to infinity of e^{5-6n} ?

The sum from n=9 to infinity of e^{5-6n} is converges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=9 to infinity of e^{5-6n}

n = 9 \sum \infty e^{5 - 6 n}

converges

Solution steps

n = 9 \sum \infty e^{5 - 6 n}

Apply Series Ratio Test:

converges

= converges

\int (\frac{1}{1 + 9 x ^{2}}) d x

\frac{d}{d x} (5 sin (- 4 x))

\int_{0}^{1} - 12 x^{2} (4 x^{3} - 1)^{3} d x

\frac{1}{1 + e ^{x^{2}}}

\int \frac{1}{25 - 36 x ^{2}} d x

What is the sum from n=9 to infinity of e^{5-6n} ?
The sum from n=9 to infinity of e^{5-6n} is converges